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Let p,q be prime.

Is it true that

$$\sum_{p\leq x}(\log p)^2 \sim x \log x~~~?\hspace{5mm}(1)$$

I haven't been able to support modest numerical evidence with the little I know of relations involving sums of logs.

The reason I wonder is that if this is true then

$$\sum_{pq \leq x}\log p\log q \sim x \log x\hspace{5mm}(2) $$ by Selberg's relation.*

I guess it would also be true that $$\sum_{p\leq n}(\log p)^2 \sim n\log n\sim p_n.$$

This has possibly been asked before but a quick search here didn't turn up anything. It may not be true!

Thanks for any insights, hints, references.

*Selberg: $\sum_{p \leq x}(\log p)^2+ \sum_{pq \leq x}\log p\log q = 2x\log x + O(x)$

daniel
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1 Answers1

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For some $\eta\in(n,n+1)$, $$ \begin{align} &\sum_{n\le x}\pi(n)(\log(n+1)^2-\log(n)^2)\\ &=\sum_{n\le x}\pi(n)\frac{2\log(\eta)}{\eta}\\ &=\sum_{n\le x}\left(\frac{n}{\log(n)}+O\left(\frac{n}{\log(n)^2}\right)\right)\left(\frac{2\log(n)}{n}+O\left(\frac{\log(n)}{n^2}\right)\right)\\ &=2x+O\left(\frac{x}{\log(x)}\right) \end{align} $$ Therefore, summing by parts, $$ \begin{align} \sum_{p\le x}\log(p)^2 &=\sum_{n\le x}\log(n)^2(\pi(n)-\pi(n-1))\\[6pt] &=\log(x)^2\pi(x)-\sum_{n\le x}\pi(n)(\log(n+1)^2-\log(n)^2)\\ &=\log(x)^2\left(\frac{x}{\log(x)}+\frac{x}{\log(x)^2}+O\left(\frac{x}{\log(x)^3}\right)\right)-2x+O\left(\frac{x}{\log(x)}\right)\\ &=x\log(x)-x+O\left(\frac{x}{\log(x)}\right) \end{align} $$

Experimental Results $$ \begin{array}{r|c|c|c} x&\sum_{p\le x}\log(p)^2&x\log(x)-x&\frac{x}{\log(x)}\\ \hline\\ 100 & 309.0926254 & 360.5170186 & 21.71472410\\ 1000 & 5686.965759 & 5907.755279 & 144.7648273\\ 10000 & 81399.38488 & 82103.40372 & 1085.736205\\ 100000 & 1048435.866 & 1051292.546 & 8685.889638 \end{array} $$

robjohn
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  • The first few lines of this are really nice and the technique seems very handy. Is there a current text containing proofs along these lines you might recommend? – daniel Jun 15 '13 at 14:00
  • @daniel: Sorry, I am pretty new to this particular type of proof, too, so I don't have any references. I used similar methods in this answer. – robjohn Jun 15 '13 at 14:08
  • For some reason I was fretting about the -x term but if $(x\ln x - x )\sim x \ln x,$ your answer would indicate the answer to the original question is actually yes...is that correct? – daniel Jun 16 '13 at 14:41
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    @daniel: Oh, indeed; my answer simply gives a bit more accuracy to the statement. $x\log(x)-x\sim x\log(x)$ since $\lim\limits_{x\to\infty}\frac{x}{x\log(x)}=0$. – robjohn Jun 16 '13 at 15:26